What are the constraints on $\alpha$ so that $AX=B$ has a solution?

Question

I found the following problem and I'm a little confused.

Consider $$A= \left( \begin{array}{ccc} 3 & 2 & -1 & 5 \\ 1 & -1 & 2 & 2\\ 0 & 5 & 7 & \alpha \end{array} \right)$$ and $$B= \left( \begin{array}{ccc} 0 & 3 \\ 0 & -1 \\ 0 & 6 \end{array} \right)$$

What are the constraints on $\alpha$ so that the matrix equation $AX=B$ has solution?

Since neither $A$ nor $B$ are square, I can't get their inverses. Is the problem wrong?

Answer 1

Ignoring the fourth column, notice that

$$\begin{pmatrix} 3 & 2 & -1 \\ 1 & -1 & 2\\ 0 & 5 & 7 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}$$

and

$$\begin{pmatrix} 3 & 2 & -1 \\ 1 & -1 & 2\\ 0 & 5 & 7 \end{pmatrix} \begin{pmatrix} \frac15 \\ \frac65 \\ 0 \end{pmatrix}=\begin{pmatrix}3 \\ -1 \\ 6 \end{pmatrix}.$$

Can you comment on whether $\alpha$ influence existence of solutions?

Are you able to construct an $X$ for the original problem?

Remark:

In general, you might like to perform row operations on the system of equations.

Tips:

If $A_1$ is a matrix that consists of some columns of $A$ and $A_1$ is non-singular, the solution to $AX=B$ always exist. Do you see why?

Answer 2

Use Gauss-Jordan elimination to find the RREF of the augmented matrix $[\mathrm A \,|\, \mathrm B]$. We have

$$\left[ \begin{array}{cccc|cc} 3 & 2 & -1 & 5 & 0 & 3\\ 1 & -1 & 2 & 2 & 0 & -1\\ 0 & 5 & 7 & \alpha & 0 & 6\end{array} \right]$$

Using SymPy,

>>> t = symbols('t')
>>> M = Matrix([[3, 2, -1, 5, 0, 3], [1, -1, 2, 2, 0, -1], [0, 5, 7, t, 0, 6]])
>>> M
⎡3  2   -1  5  0  3 ⎤
⎢                   ⎥
⎢1  -1  2   2  0  -1⎥
⎢                   ⎥
⎣0  5   7   t  0  6 ⎦
>>> M.rref()
⎛⎡           3⋅t   123        ⎤           ⎞
⎜⎢1  0  0  - ─── + ───  0  1/5⎥, [0, 1, 2]⎟
⎜⎢            70    70        ⎥           ⎟
⎜⎢                            ⎥           ⎟
⎜⎢           t    1           ⎥           ⎟
⎜⎢0  1  0    ── - ──    0  6/5⎥           ⎟
⎜⎢           10   10          ⎥           ⎟
⎜⎢                            ⎥           ⎟
⎜⎢           t    1           ⎥           ⎟
⎜⎢0  0  1    ── + ──    0   0 ⎥           ⎟
⎝⎣           14   14          ⎦           ⎠

We can conclude that the augmented matrix has full row rank. What does that imply?